THE GOLINSKII-IBRAGIMOV METHOD AND A THEOREM OF DAMANIK AND KILLIP

BARRY SIMON

Abstract. In 1971, Golinskii and Ibragimov proved that if the Verblunsky coefficients, {α_{n}}_{n=0}^{∞}, of a measure dμ on ∂𝔻 obey ∑_{n=0}^{∞} n|α_{n}|^{2} < ∞, then the singular part, dμ_{s}, of dμ vanishes. We show how to use extensions of their ideas to discuss various cases where ∑_{n=0}^{N} n|α_{n}|^{2} diverges logarithmically. As an application, we provide an alternative to a part of the proof of a recent theorem of Damanik and Killip.

1. Introduction

Let ∂𝔻 be the unit circle {z | |z| = 1} in ℂ and 𝔻 the open disc, {z | |z| < 1}. Let μ be a probability measure on ∂𝔻 which is not supported on a finite number of points. Then using the Gram-Schmidt procedure, we can define monic orthogonal polynomials on the unit circle (OPUC) Φ_{n}(z; dμ) and normalized polynomials φ_{n}(z; dμ). These obey the Szegő recursion formulae [8, 9]:

Φ_{n+1}(z) = zΦ_{n}(z) − ᾱ_{n}Φ_{n}^{∗}(z)  (1.1)

where

Φ_{n}^{∗}(z) = z^{n} \overline{Φ_{n}(\frac{1}{z̄})}  (1.2)

The parameters α_{n} are called the Verblunsky coefficients of dμ (also called Schur, Szegő, Geronimus, or reflection parameters or coefficients). They lie in 𝔻 and any α ∈ ×_{n=0}^{∞} 𝔻 is the Verblunsky coefficient of a unique measure [2, 8]. In this note, we are mainly interested in the spectral problem of going from information on the Verblunsky coefficients to information on the measure.

Our starting point is a method from a lovely 1971 paper of B. Golinskii and I. Ibragimov [4], who used this method to prove:

Theorem 1 (Golinskii-Ibragimov [4]). If ∑_{n=0}^{∞} n|α_{n}|^{2} < ∞, then dμ_{s} = 0.

Date: March 13, 2003.

1 Mathematics 253-37, California Institute of Technology, Pasadena, CA 91125.

E-mail: bsimon@caltech.edu.

2 Supported in part by NSF grant DMS-0140592.

Here we make a canonical decomposition:

dμ = w(θ) \frac{dθ}{2π} + dμ_{s}  (1.3)

where dμ_{s} is singular with respect to dθ. We are interested in α’s that obey a weaker condition:

∑_{n=0}^{N} n|α_{n}|^{2} ≤ A log N + C  (1.4)

with A, C constant. Think of α_{n} = \sqrt{A}/n as a prototypical example. First, we will prove

Theorem 2. If (1.4) holds with A < \frac{1}{4}, then dμ_{s} = 0.

The GI method most directly only gets A < \frac{1}{16}, but by replacing their L^{1} methods by an L^{2} method, we will bring things up to A < \frac{1}{4}.

Theorem 2 is almost optimal in that there are examples with A > \frac{1}{4} but |A − \frac{1}{4}| arbitrarily small, where dμ has an eigenvalue. However, we can do better if we assume the α’s are real:

Theorem 3. If all α_{n} are real (equivalently, dμ is invariant under complex conjugation) and (1.4) holds with A < \frac{1}{2}, then dμ_{s} can only consist of possible pure points at z = +1 or z = −1. If A = \frac{1}{4}, dμ_{s} = 0.

We will use an idea motivated by Damanik-Killip [1] to prove Theorem 3. In this regard, we will prove the following special-looking result which, as we will explain, is related to [1]:

Theorem 4. Let all α_{n} be real and obey (i)

∑_{j=0}^{N} j|α_{2j}|^{2} ≤ A log N + C  (1.5)

(ii) |α_{2j−1}| ≤ |α_{2j+1}| for j = 1, 2, … with either all α_{2j−1} ≥ 0 or all

α_{2j−1} ≤ 0. (iii) ∑_{j=1}^{∞}|α_{2j−1}|^{2} < ∞

If A < \frac{1}{2}, dμ_{s} consists only of possible pure points at z = ±1 or z = ±i. Moreover, if A = \frac{1}{4}, the only possible pure points are at z = ±1, and if A = \frac{1}{4} and α_{2j−1} ≤ 0, then dμ_{s} = 0.

This theorem is custom-made to provide part of a proof of the following recent striking result of Damanik-Killip [1]:

Theorem 5 (Damanik-Killip [1]). Let H be a half-line Schrödinger operator on ℓ^{2}(ℤ_{+}),

(Hu)_{n} = u_{n+1} + u_{n−1} + v_{n}u_{n}

(u_{0} interpreted as 0). Suppose spec(H) = [−2, 2]. Then H has purely a.c. spectrum, that is, σ_{sc} = σ_{pp} = ∅.

Their proof has the following steps (they use γ_{n} for our α_{n}):

(i) Following Szegő [9], map H to an associated measure on ∂𝔻 by using z = e^{iθ} ↦ E = 2 cos θ to pull back the spectral measure dρ on [−2, 2] for H to a measure μ on ∂𝔻. Note (following Geronimus) that the Verblunsky coefficients for dμ obey:

v_{n+1} = (1 − α_{2n−1})α_{2n} − (1 + α_{2n−1})α_{2n−2}  (1.6)
1 = (1 − α_{2n−1})(1 − α_{2n}^{2})(1 + α_{2n+1})  (1.7)

(for general Jacobi matrices, the left side of (1.7) is a_{n+1}^{2}; the initial conditions are α_{−1} = −1).

(ii) Analyze (1.7) with α_{n} real and |α_{n}| < 1 to conclude: (a) [Their Lemma 4.1] α_{2n−1} ≤ α_{2n+1} ≤ 0

(b) [Their Lemma 4.2] |α_{2n+1}| ≤ \frac{1}{n+2} so ∑_{n=1}^{∞}|α_{2n−1}|^{2} < ∞.

(c) [Their Proposition 4.5] ∑_{j=0}^{N}(j + 1)α_{j}^{2} ≤ \frac{1}{4} log N + C

(iii) Translate information on the α_{n}’s to information on the v_{n}’s. (iv) Prove that −2, 2, 0 are not eigenvalues of H.

(v) Prove that solutions of

u_{n+1} + u_{n−1} + v_{n}u_{n} = Eu  (1.8)

with E ∈ (−2, 0) ∪ (0, 2) have |u_{n}| ≤ cn^{η} for any η > \frac{1}{2\sqrt{2}}. (vi) Prove that the set of E in (−2, 2) for which (1.8) has unbounded

solutions has Hausdorff dimension 0. (vii) Use (v), (vi), Hausdorff dimension, and the Jitomirskaya-Last in-

equalities [6] to show that dμ has no singular part on (−2, 0)∪(0, 2). Given step (ii), one can use Theorem 4 to replace steps (iii)–(vii).

For Theorem 4 says dμ is purely a.c. and the pull back then implies dρ is. Since Theorem 4 depends on ideas closely related to steps (iv) and (v), what we are really doing is using an appeal to the GI method to replace steps (vi) and (viii) and, in particular, the use of Hausdorff dimension and the Jitomirskaya-Last inequalities. These steps follow ideas of Remling [7] so, in essence, where [1] extends [7], we extend [4]. It is pointless to argue which approach is “simpler” (since some of their techniques have appeared extensively in the Schrödinger operator literature), but we believe it useful to have the alternate approach.

In Section 2, we discuss the ideas of Golinskii-Ibragimov and, in particular, prove Theorem 2. In Section 3, we use Prüfer variables for OPUC to prove Theorem 3. In Section 4, we prove Theorem 4.

It is a pleasure to thank David Damanik and Rowan Killip for telling me of their work and for useful discussions.

2. The GI Method

Golinskii-Ibragimov make the assumption that

∑_{n=0}^{∞} n|α_{n}|^{2} < ∞  (2.1)

On ∂𝔻, D^{−1} is defined a.e. dθ as boundary values of D, the Szegő function (see [3, 8]), and so a.e. dμ_{ac}. We extend it to L^{2}(∂𝔻, dμ) = L^{2}(∂𝔻, dμ_{ac}) ⊕ L^{2}(∂𝔻, dμ_{s}) by setting to 0 on the singular subspace. Then Golinskii-Ibragimov [4] prove that

∥φ_{n}^{∗} − D^{−1}∥_{L^{2}(∂𝔻,dμ)} ≤ Cn^{−1/2}  (2.2)

and that

∥(φ_{n}^{∗})^{−1}∥_{∞} ≤ C_{1} exp(C_{2}\sqrt{log n})  (2.3)

where ∥ · ∥_{∞} means sup over ∂𝔻 or 𝔻̄ (equal since (φ_{n}^{∗})^{−1} is analytic in a neighborhood of 𝔻̄). In (2.2), the Szegő function exists since (2.1) implies ∑_{n=0}^{∞}|α_{n}|^{2} < ∞. We will actually prove results like this below.

They then write

∥|φ_{n}^{∗}|^{−2} − |D|^{2}∥_{L^{1}(∂𝔻,dθ/2π)}
≤ ∥(φ_{n}^{∗})^{−1}∥_{∞}^{2}∥(|φ_{n}|^{2} − |D|^{−2})D^{2}∥_{L^{1}(∂𝔻,dθ/2π)}
≤ ∥(φ_{n}^{∗})^{−1}∥_{∞}^{2}∥(|φ_{n}|^{2} − |D|^{−2})∥_{L^{1}(∂𝔻;dμ)}  (2.4)
≤ ∥(φ_{n}^{∗})^{−1}∥_{∞}^{2}∥φ_{n} − D^{−1}∥_{L^{2}(∂𝔻;dμ)}∥|φ_{n}| + |D|^{−1}∥_{L^{2}(∂𝔻;dμ)} (2.5)

In (2.4), we use D^{2} \frac{dθ}{2π} = dμ_{ac} ≤ dμ and in (2.5) that ||φ_{n}| − |D|^{−1}| ≤ |φ_{n} − D^{−1}|. Thus, by (2.2) and (2.3), |φ_{n}|^{−2} \frac{dθ}{2π} → |D|^{2} \frac{dθ}{2π} in norm on measures. But it is known that |φ_{n}|^{−2} \frac{dθ}{2π} → dμ weakly (see, e.g., [2]). Thus dμ = |D|^{2} \frac{dθ}{2π} so dμ_{s} = 0.

What is especially interesting about this approach is that it uses the divergent estimate (2.3). Clearly, we can have much more rapid growth of ∥(φ_{n}^{∗})^{−1}∥_{∞} than in (2.3) and still have convergence. Basically, it suffices to have ∥(φ_{n}^{∗})^{−1}∥_{∞} ≤ Cn^{β}, with 2β < \frac{1}{2}. If one looks at the proof of (2.3) in [4], that translates to a bound like (1.4) with \sqrt{A} < \frac{1}{4}, that is, A < \frac{1}{16}. Our first observation is that instead of estimating L^{1} norms as GI do, it pays to estimate L^{2} norms. ∥(φ_{n}^{∗})^{−1}∥_{∞} will then occur as a first power, not second. Here are the key facts:

Theorem 2.1. Let μ obey the Szegő condition. Then

(a) dμ_{s} = 0 if and only if ∥(φ_{n}^{∗})^{−1} − D∥_{L^{2}(∂𝔻,dθ/2π)} → 0.

(b) If I is an open interval and ∥χ_{I}[(φ_{n}^{∗})^{−1} − D]∥_{L^{2}(∂𝔻,dθ/2π)} → 0, then μ_{s}(I) = 0.

Proof. (a) As is well known (see [2]), ∥(φ_{n}^{∗})^{−1}∥_{L^{2}(∂𝔻,dθ/2π)} = 1. Moreover, (φ_{n}^{∗})^{−1}(z) → D(z) uniformly on compact subsets of 𝔻, so (φ_{n}^{∗})^{−1} → D weakly in L^{2}. Thus (φ_{n}^{∗})^{−1} → D in norm if and only if 1 = ∥(φ_{n})^{−1}∥_{L^{2}(∂𝔻,dθ/2π)}^{2} = ∥D∥_{L^{2}(∂𝔻,dθ/2π)}^{2}. But ∥D∥^{2} = ∫ |D(θ)|^{2} \frac{dθ}{2π} = μ_{ac}(∂𝔻), so norm convergence is equivalent to μ_{s}(∂𝔻) = 0.

(b) Let f be continuous and have support in I. Then

f|φ_{n}^{∗}|^{−2} = (f(φ_{n}^{∗})^{−1}χ_{1})( \overline{(φ_{n}^{∗})^{−1}} χ_{I})
→ f|D|^{2}

in L^{1} and thus, by the weak convergence of |φ_{n}^{∗}|^{−2} \frac{dθ}{2π} to dμ, we have ∫ fdμ = ∫ f|D|^{2} \frac{dθ}{2π}, that is, dμ_{s}(I) = 0. □

Theorem 2.2. If

∥χ_{I}(φ_{n}^{∗})^{−1}∥_{∞}∥(φ_{n}^{∗} − D^{−1})∥_{L^{2}(∂𝔻,dμ)} → 0  (2.6)

for some open interval I (including I = ∂𝔻), then μ_{s}(I) = 0.

Proof.

∥χ_{I}[(φ_{n}^{∗})^{−1} − D]∥_{L^{2}(∂𝔻,dθ/2π)} = ∥χ_{I}(φ_{n}^{∗})^{−1}(φ_{n}^{∗} − D^{−1})D∥_{L^{2}(∂𝔻,dθ/2π)}
≤ ∥χ_{I}(φ_{n}^{∗})^{−1}∥_{∞}∥(φ_{n}^{∗} − D^{−1})∥_{L^{2}(∂𝔻,dμ_{ac})}
≤ LHS of (2.6)

Thus the result follows from Theorem 2.1. □

Define ρ_{n} = (1 − |α_{n}|^{2})^{1/2}. A straightforward calculation shows that for n < m,

⟨φ_{n}^{∗}, φ_{m}^{∗}⟩ = ∏_{ℓ=n}^{m−1} ρ_{ℓ}

and, of course, ⟨φ_{n}^{∗}, φ_{n}^{∗}⟩ = 1. Since [3] φ_{n}^{∗} → D^{−1} in L^{2}(∂𝔻, dμ), we have that

∥φ_{n}^{∗} − D^{−1}∥_{L^{2}(∂𝔻,dμ)}^{2} = 2(1 − ∏_{ℓ=n}^{∞} ρ_{ℓ})  (2.7)

Proposition 2.3. (a) ∥φ_{n}^{∗} − D^{−1}∥_{L^{2}(∂𝔻,dμ)}^{2} ≤ 2 ∑_{ℓ=n}^{∞}|α_{ℓ}|^{2}

(b) If (1.4) holds, then ∑_{n=0}^{∞}|α_{n}|^{2} < ∞.

(c) If (1.4) holds for any A, then

∥φ_{n}^{∗} − D^{−1}∥_{L^{2}(∂𝔻,dμ)} ≤ \frac{C log n}{n^{1/2}}  (2.8)

for all large n.

Proof. (a) 1 − ∏_{ℓ=n}^{∞} ρ_{ℓ} ≤ 1 − ∏_{ℓ=n}^{∞} ρ_{ℓ}^{2} since ρ_{ℓ}^{2} ≤ ρ_{ℓ}. Moreover, 1 − ∏_{ȷ=1}^{ℓ} x_{j} ≤ ∑_{j=1}^{ℓ}(1 − x_{j}) by a simple induction. Thus (2.3) implies (a).

(b),(c)

∑_{n=2^{ℓ}}^{2^{ℓ+1}−1} |α_{n}|^{2} ≤ 2^{−ℓ} ∑_{n=0}^{2^{ℓ+1}} n|α_{n}|^{2}
≤ 2^{−ℓ}[A log(2^{ℓ+1}) + C]

Since

∑_{ℓ=k}^{∞} ℓ2^{−ℓ} ≤ Ck2^{−k}

we see that (1.4) implies ∑_{n=0}^{∞}|α_{n}|^{2} < ∞ and

∑_{ℓ=n}^{∞} |α_{ℓ}|^{2} ≤ Cn^{−1} log n

Given (a), we get (c). □

Theorem 2.4. If the α’s obey (1.4) and for some open interval I ⊂ ∂𝔻, and B < \frac{1}{2}, we have

sup_{z∈I} [|φ_{n}(z)|^{−1}] ≤ C(n + 1)^{B}  (2.9)

for a constant C, then μ_{s}(I) = 0.

Proof. For z ∈ ∂𝔻, |φ_{n}^{∗}(z)| = |φ_{n}(z)| so (2.9) says ∥χ_{I}(φ_{n}^{∗})^{−1}∥_{∞} ≤ C(n + 1)^{B}. This and (2.8) implies that (2.6) holds, so μ_{s}(I) = 0. □

The second kind polynomials, ψ_{n}, Ψ_{n}, are defined by reversing the signs of all the Verblunsky coefficients. It is known (see, e.g., [5]) that for z ∈ ∂𝔻, Re(φ_{n}(z)ψ̄_{n}(z)) = 1, so |φ_{n}(z)|^{−1} ≤ |ψ_{n}(z)|.

Since ψ_{n} = (∏_{j=0}^{n−1} ρ_{j}^{−1})Ψ_{n} and ∑ α_{j}^{2} < ∞ implies ∏_{j=0}^{∞} ρ_{j}^{−1} < ∞, we see |ψ_{n}(z)| ≤ C|Ψ_{n}|. Thus, Theorem 2.4 can be rewritten:

Corollary 2.5. If the α’s obey (1.4) and for some open interval I ⊂ ∂𝔻, we have that

sup_{z∈I} |Ψ_{n}(z)| ≤ C(n + 1)^{B}

for a constant C, and B < \frac{1}{2}, then μ_{s}(I) = 0.

Proof of Theorem 2. (1.2) implies |Ψ_{n}^{∗}| = |Ψ_{n}| on ∂𝔻 and so (1.1) implies |Ψ_{n+1}| ≤ (1 + |α_{n}|)|Ψ_{n}|. Thus

sup_{z∈∂𝔻}|Ψ_{n}(z)| ≤ ∏_{j=0}^{n−1} 1 + |α_{j}| ≤ exp( ∑_{j=0}^{n−1} |α_{j}|)  (2.10)

But, by the Schwartz inequality, (C is a “variable constant” and ε arbitrarily small)

∑_{j=0}^{n} |α_{j}| ≤ ( ∑_{j=0}^{n}(j + 1)|α_{j}|^{2})^{1/2}( ∑_{j=0}^{n}(j + 1)^{−1})^{1/2}
≤ [A log(n) + C]^{1/2} [log(n) + C]^{1/2}  (2.11)
≤ \sqrt{A + ε} log(n) + C

Thus, by (2.10),

sup_{z∈𝔻}|Ψ_{n}(z)| ≤ Cn^{\sqrt{A+ε}}  (2.12)

If \sqrt{A + ε} < \frac{1}{2}, that is, A < \frac{1}{4}, Corollary 2.5 is applicable. □

We emphasize that, in essence, the calculation in Proposition 2.3(a), (2.10), and the basic strategy are all from [4]; the only real advance in this section is the use of L^{2}-norms allowing A < \frac{1}{4}, where the method of [4] gets A < \frac{1}{16}.

Finally, we note A = \frac{1}{4} is a critical value for the appearance of bound states, for

Theorem 2.6. If all α_{n} are real, then φ_{n}(1) > 0 and

φ_{n}(1) = ∏_{j=0}^{n−1}(1 − α_{j})  (2.13)

In addition, (−1)^{n}φ_{n}(−1) > 0 and

(−1)^{n}φ_{n}(−1) = ∏_{j=0}^{n−1}(1 − (−1)^{j+1}α_{j})  (2.14)

Remark. In particular, if α_{j} = B(j + 1)^{−1}, then φ_{n}(1) ∼ n^{−B} and ∑|φ_{n}(1)|^{2} < ∞ if B > \frac{1}{2}. For that α, (1.4) holds for A = B^{2}, that is, there are examples with dμ_{s} ≠ 0 and (1.4) holding for any A > \frac{1}{4}. At A = \frac{1}{4}, our proof of Theorem 2 shows |α_{n}(z)| ≥ Cn^{−1/2}, so at least there are no eigenvalues.

Proof. φ_{n}(1) is real by induction and then (1.1) says

φ_{n+1}(1) = (1 − α_{n})φ_{n}(1)

proving (2.13). Similarly, φ_{n}(−1) is real. (1.1) for z = −1 says

(−1)^{n+1}φ_{n+1}(−1) = (−1)^{n}φ_{n}(−1) − α_{n}(−1)^{n+1}(−1)^{n}φ_{n}(−1)

since φ_{n}^{∗}(−1) = (−1)^{n}φ_{n}(−1), which yields (2.14). □

3. Prüfer Variables and the Proof of Theorem 3

Write Ψ_{n}(z) = R_{n}(z)e^{i[θ_{n}(z)+nη(z)]} where e^{iη(z)} = z and θ_{n} is defined initially only modulo 2π. If α_{n} is real, (1.1) with α_{n} → −α_{n} becomes R_{n+1}^{2} = R_{n}^{2}|e^{iη}e^{i[θ_{n}+nη]} + α_{n}e^{−iθ_{n}}|^{2}.

Thus,

\frac{R_{n+1}^{2}}{R_{n}^{2}} = 1 + α_{n}^{2} + 2α_{n} cos((n + 1)η + 2θ_{n})  (3.1)

Secondly,

e^{i(θ_{n+1}(z)−θ_{n}(z))} = \frac{1 + α_{n}e^{−i(2θ_{n}+(n+1)θ_{n})}}{[1 + α_{n}^{2} + 2α_{n} cos((n + 1)η + 2θ_{n})]^{1/2}}  (3.2)

These are the Prüfer variable equations for Ψ. (3.2) implies cos(θ_{n+1} − θ_{n}) > 0, so we can always pick θ_{n+1} so |θ_{n+1} − θ_{n}| < \frac{π}{2}, settling the 2π ambiguity.

Proposition 3.1. Let α_{n} be real and ∑ α_{n}^{2} < ∞. Define Q = sup_{n}|α_{n}| < 1. Then

(a) For a z independent constant C ∈ (1, ∞),

C^{−1} ≤ R_{n} exp(− ∑_{j=0}^{n−1} α_{j} cos((j + 1)η + 2θ_{j}(z))) ≤ C  (3.3)

(b) For all n,

|θ_{n+1} − θ_{n}| ≤ \frac{π}{2} \frac{1}{1−Q} |α_{n}|  (3.4)

Remark. C depends only on Q and ∑_{j=0}^{∞} α_{j}^{2}.

Proof. (a) Define b_{n}(z) so the right side of (3.1) is 1 + b_{n}. Then (1 − Q)^{2} ≤ (1 − |α_{n}|)^{2} ≤ 1 + b_{n} ≤ (1 + α_{n})^{2} ≤ (1 + Q)^{2}. It follows that for a Q dependent constant K, we have e^{−Kb_{n}^{2}} ≤ (1 + b_{n})e^{−b_{n}} ≤ e^{Kb_{n}^{2}}. Since b_{n} ≤ 3|α_{n}| and b_{n} has an α_{n}^{2} in it, we have

exp(−(9K+1)α_{n}^{2}) ≤ (1+b_{n}) exp(−2α_{n} cos((n+1)η+2θ_{n})) ≤ exp((9K+1)α_{n}^{2})

Thus, by (3.1) and R_{0} = 1, we have (3.3) with

C = exp((9K + 1) ∑_{j=0}^{∞} α_{j}^{2})

(b) Taking imaginary parts of both sides of (3.2) and using the lower bound 1 − Q on the denominator, we get

|sin(θ_{n+1} − θ_{n})| ≤ |α_{n}|(1 − Q)^{−1}  (3.5)

(3.2) also implies cos(θ_{n+1} − θ_{n}) > \frac{(1−Q)}{(1+Q)} > 0, so |θ_{n+1} − θ_{n}| < \frac{π}{2}. Since |x_{n}| < \frac{π}{2} implies |x| ≤ \frac{π}{2}|sin x|, (3.4) follows from (3.5). □

The point of (3.3) is to control |R_{n}|, we need to control ∑_{j=0}^{n−1} α_{j} cos((j + 1)η + 2θ_{j}). In using the Schwartz inequality, we will decouple the α_{j}’s and the cosines so the key will be the following (essentially in [1]):

Lemma 3.2. (i) If k is not a multiple of 2π,

| ∑_{j=1}^{n} \frac{cos(kj + θ_{j})}{j}| ≤ \frac{1}{|sin(\frac{k}{2})|} [1 + ∑_{j=1}^{n−1} \frac{|θ_{j+1} − θ_{j}|}{j + 1}]  (3.6)

(ii) If k is not a multiple of π,

∑_{j=1}^{N} \frac{cos^{2}(kj + θ_{j})}{j} ≤ \frac{1}{2} (log N + 1 + C)  (3.7)

where

C = \frac{1}{|sin(k)|} [1 + 2 ∑_{j=1}^{∞} \frac{|θ_{j+1} − θ_{j}|}{j + 1}]  (3.8)

Proof. Since cos(kj + θ_{j}) = Re(exp(ikj) exp(iθ_{j})), it suffices to prove (3.6) with cosines replaced by complex exponentials . Define b_{n} = ∑_{j=1}^{n} e^{ikj} so, by summing the geometric series,

|b_{n}| ≤ \frac{1}{|sin(\frac{k}{2})|}  (3.9)

If

a_{j} = \frac{e^{iθ_{j}}}{j}

then the sum we want to control is ∑_{j=1}^{n}(b_{j} − b_{j−1})a_{j} with b_{0} = 0. But

∑_{j=1}^{n}(b_{j} − b_{j−1})a_{j} = − ∑_{j=1}^{n} b_{j}(a_{j+1} − a_{j})

where a_{n+1} = 0. Thus

| ∑_{j=1}^{n} \frac{cos(kj + θ_{j})}{j}| ≤ \frac{1}{sin(\frac{k}{2})} [ ∑_{j=1}^{n−1} |a_{j+1} − a_{j}| + |a_{n}|]  (3.10)

Clearly,

|a_{j+1} − a_{j}| ≤ |(e^{iθ_{j+1}} − e^{iθ_{j}})(j + 1)^{−1}| + |e^{iθ_{j}}(j^{−1} − (j + 1)^{−1})|
≤ |θ_{j+1} − θ_{j}|(j + 1)^{−1} + j^{−1} − (j + 1)^{−1}  (3.11)

Since ∑_{j=1}^{n−1} j^{−1} − (j + 1)^{−1} + n^{−1} = 1, (3.10) and (3.11) yield (3.6).

(b) Since cos^{2}(x) = \frac{1}{2}(1 + cos(2x)) and

∑_{j=1}^{n} \frac{1}{j} ≤ 1 + ∫_{1}^{n} \frac{dx}{x} = 1 + log(n)

(3.6) implies (3.7). □

Proof of Theorem 3. Write

\begin{matrix}| ∑_{j=0}^{n−1} & α_{j} cos((j + 1)η + 2θ_{j})| \\ ≤ ( ∑_{j=0}^{n−1}(j + 1)|α_{j}|^{2}))^{1/2}( ∑_{j=0}^{n−1} \frac{cos^{2}((j + 1)η + 2θ_{j})}{(j + 1)})^{1/2} \end{matrix} (3.12)

By hypothesis, the first sum on the right side of (3.12) is bounded by A log N + C. By the lemma, if η is not a multiple of π (i.e., z ≠ ±1), the second sum is bounded by

\frac{1}{2} (log n + C̃|sin(η)|^{−1})

where C̃ can be chosen independently of z, since

∑_{j=1}^{∞} \frac{|θ_{j+1} − θ_{j}|}{j + 1} ≤ C ∑_{j=0}^{∞} \frac{|α_{j}|}{j + 2}  (by (3.5))

≤ C( ∑_{j=0}^{∞}|α_{j}|^{2})^{1/2}( ∑_{j=0}^{∞} \frac{1}{(j + 2)^{2}} )^{1/2}

is finite. Thus, for η ∈ [θ_{0}, π − θ_{0}],

|R_{n}| ≤ C exp|[A(log n) + C]|^{1/2}[\frac{1}{2} log n + C̃|sin θ_{0}|^{−1}]^{1/2}
≤ C_{ε}n^{\sqrt{\frac{1}{2} A+ε}}

where C_{ε} depends on ε and θ_{0}. So long as \sqrt{\frac{1}{2}A} < \frac{1}{2}, we can apply Corollary 2.5, that is, A < \frac{1}{2}. We conclude dμ_{s}(I) = 0 for I = ±(θ_{0}, π − θ_{0}), that is, dμ_{s} is supported on {±1}. As already noted, at ±1, |φ_{n}(±1)| ≥ n^{−\sqrt{A}}, so if A = \frac{1}{4}, φ_{n}(±1) are not in L^{2} and dμ_{s} = 0. □

4. Sequences of Bounded Variation and the Proof of Theorem 4

To obtain Theorem 4, we need one more summation-by-parts argument that will supplement Lemma 3.2:

Lemma 4.1. If k is not a multiple of 2π,

| ∑_{j=1}^{n}c_{j} cos(kj + θ_{j})|

≤ \frac{1}{sin(\frac{k}{2})}[ ∑_{j=1}^{∞} |c_{j+1} − c_{j}| + sup_{n}|c_{n}| + ∑_{j=1}^{∞} |c_{j+1}| |θ_{j+1} − θ_{j}|]

(4.1)

Proof. As in the proof of Lemma 3.2, let b_{n} = ∑_{j=1}^{n} e^{ikj}, so (3.9) holds and a_{k} = c_{j}e^{iθ_{j}}. Then summing by parts as in the earlier lemma,

| ∑_{j=1}^{n} c_{j} cos(kj + θ_{j})| ≤ \frac{1}{sin(\frac{k}{2})}[|c_{n}| + ∑_{j=1}^{n−1} |a_{j+1} − a_{j}|]

But

|a_{j+1} − a_{j}| ≤ |a_{j+1}| |e^{iθ_{j+1}} − e^{iθ_{j}}| + |c_{j+1} − c_{j}|

so (4.1) follows. □

Proof of Theorem 4. By (3.3) and the hypothesis that ∑_{j=0}^{∞}|α_{j}|^{2},

|Ψ_{n}(z)| ≤ C exp( ∑_{j=0}^{n−1} α_{j} cos((j + 1)η + 2θ_{j}))

Write

∑_{j=0}^{n−1} α_{j} cos((j + 1)η + 2θ_{j}) = O_{n} + E_{n}  (4.2)

where O_{n} is the sum over odd values of j and E_{n} over even values. By Lemma 4.1,

\begin{aligned}|O_{n}| ≤ \frac{1}{|sin(\frac{η}{2})|}[ sup_{n}|α_{2n−1}| & + ∑_{n=1}^{∞}|α_{2n+1} − α_{2n−1}| \\ & + ∑_{n=1}^{∞}|α_{2n+1}| |θ_{2n+1} − θ_{2n−1}|] \end{aligned}  (4.3)

Since α_{2n−1} is monotone, ∑_{n=1}^{∞}|α_{2n+1}−α_{2n−1}| = |α_{1}|. By (3.4), |θ_{2n+1}− θ_{2n−1}| ≤ C(|α_{2n−1}| + |α_{2n}|) so, since ∑_{n=0}^{∞}|α_{n}|^{2} < ∞, (4.3) implies

|O_{n}| ≤ C|sin(\frac{η}{2})|^{−1}  (4.4)

For E_{n}, we use the same argument as in the proof of Theorem 3 taking into account that the change in frequency from 2n to 2n + 2 is 2η + (θ_{2n+2} − θ_{2n}). Thus |sin(η)|^{−1} becomes |sin(2η)|^{−1}, and we find that

|E_{n}| ≤ (A log N + C)^{1/2}(\frac{1}{2} log N + C|sin(2η)|^{−1})^{1/2}  (4.5)

(4.4) and (4.5) imply that for any θ_{0} > 0 for all z ∈ (θ_{0}, \frac{π}{2} − θ_{0}) ∪ (\frac{π}{2} + θ_{0}, π − θ_{0}),

|Ψ_{n}| ≤ C_{θ_{0},ε}n^{ε+\sqrt{A/2}}

which, by Corollary 2.5, implies μ_{s} is restricted to ±1, ±i.

To obtain the result on eigenvalues when A = \frac{1}{4}, note first that since |E_{n}| ≤ ∑_{m=1}^{[n/2]}|α_{2m}|, uniformly in z,

e^{|E_{n}|} ≤ Cn^{1/2}

if A = \frac{1}{4}. At z = ±i, (4.4) implies e^{|O_{n}|} is bounded, so |φ_{n}(±i)| ≥ Cn^{−1/2} is not in L^{2}.

At ±1, we use Theorem 2.6. Since α_{2j−1} ≤ 0, (2.13) implies

|φ_{n}(±1)| ≥ ∏_{\begin{matrix}j=0 \\ j even \end{matrix}}^{n}(1 − |α_{j}|) ≥ Cn^{−1/2}

since 1 − α_{2j−1} ≥ 1. So ±1 are not eigenvalues if α_{2j−1} ≤ 0. □

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